Integration by Partial Fractions Exercises

Example 1

First we have to put the fractions over a common denominator by multiplying each fraction by a clever form of 1. After the fractions have the same denominator, we can add their numerators.

If there was any part of that problem that gave you trouble, go review fraction addition now. If it was a piece of cake, go ahead and try the next one.

Example 2

Find

Answer

As with the numerical fractions, we have to put the rational functions over a common denominator by multiplying each by a clever form of 1. Once the denominators are the same, we can add the numerators.

We could continue to the next step, which would be

but we're not going to bother. We have the right form for the partial fractions technique when we get to the step

so we'll stop there.

When adding rational functions for the sake of the partial fractions technique, we don't multiply things out. For example, when adding

we stop here:

We also don't bother to write out the step where we multiply each fraction by a clever form of 1. We just multiply the numerator of each fraction by the denominator of the other fraction, then add the results.

If this doesn't make sense, we recommend reviewing how to add rational functions. If this does make sense, it's time to practice.

Example 3

Find the sum. A and B are unknown numbers.

Answer

Example 4

Decompose into partial fractions.

Answer

We need to find A and B so that

Adding the partial fractions,

We must have the numerators equal, so we need to find A and B such that

10x + 27 = A(x + 3) + B(x + 2).

Take x = -3 so we can find B without worrying about A:

10x + 27 = A(x + 3) + B(x + 2)

10(-3) + 27 = A((-3) + 3) + B((-3) + 2)

-3 = -B

B = 3.

Now take x = 0 and find A:

10x + 27 = A(x + 3) + B(x + 2)

10(0) + 27 = A((0) + 3) + (3)((0) + 2)

27 = 3A + 6

7 = A

The final decomposition is

Example 5

Decompose into partial fractions.

Answer

We need to find A and B so that

This means we want

and so the numerators must be equal:

8x – 3 = A(4x – 5) + B(2x + 1).

Take (because that's a little nicer than ) and solve for A:

Then take x = 0 and solve for B:

We get the decomposition

Example 6

Decompose into partial fractions.

Answer

We need to find A and B so that

Add the partial fractions and set the resulting numerator equal to the numerator of the original rational function:

11 – 23x = A(2 – 5x) + B(1 – x).

Take x = 1 to solve for A:

11 – 23x = A(2 – 5x) + B(1 – x)

11 – 23(1) = A(2 – 5(1)) + B(1 – 1)

-12 = -3A

4 = A

Then take x = 0 and solve for B:

11 – 23x = A(2 – 5x) + B(1 – x)

11 – 23(0) = (4)(2 – 5(0)) + B(1 – 0)

11 = 8 + B

3 = B

We get the decomposition

Example 7

Decompose into partial fractions.

Answer

We need to find A and B so that

Add the partial fractions and set the resulting numerator equal to the numerator of the original function:

3 – 10x = A(1 – x) + B(3x + 4).

Take x = 1 and solve for B:

3 – 10x = A(1 – x) + B(3x + 4)

3 – 10(1) = A(1 – (1)) + B(3(1) + 4)

-7 = 7B

-1 = B

Then take x = 0 and solve for A:

3 – 10x = A(1 – x) + B(3x + 4)

3 – 10(0) = A(1 – (0)) + (-1)(3(0) + 4)

3 = A – 4

7 = A

Instead of writing the decomposition as

we write it as

Example 8

Decompose into partial fractions.

Answer

We need to find A and B so that

Add the partial fractions and set the resulting numerator equal to the numerator of the original function:

2x + 4 = A(x + 1) + B(2x + 1).

Take x = -1 and solve for B:

2x + 4 = A(x + 1) + B(2x + 1)

2(-1) + 4 = A(-1 + 1) + B(2(-1) + 1)

2 = -B

-2 = B.

Then take x = 0 and solve for A:

2x + 4 = A(x + 1) + B(2x + 1)

2(0) + 4 = A((0) + 1) + (-2)(2(0) + 1)

4 = A – 2

6 = A

We write the decomposition

Example 9

Decompose into partial fractions.

Answer

The denominator factors as

x2 + 3x + 2 = (x + 1)(x + 2)

so the partial fraction decomposition will look like

We set the original numerator and the numerator of the sum of the partial fractions equal to each other:

2x + 3 = A(x + 2) + B(x + 1).

We set x = -1 and solve for A:

2x + 3 = A(x + 2) + B(x + 1)

2(-1) + 3 = A(-1 + 2) + B(-1 + 1)

1 = A

We set x = 0 and solve for B:

2x + 3 = A(x + 2) + B(x + 1)

2(0) + 3 = (1)(0 + 2) + B(0 + 1)

3 = 2 + B

1 = B

The decomposition is

Example 10

Decompose into partial fractions.

Answer

The denominator factors as

2x2 – 5x – 3 = (x – 3)(2x + 1)

so the partial fraction decomposition will look like

Add the partial fractions and set their numerator equal to the original numerator:

8x – 10 = A(2x + 1) + B(x – 3).

Set x = 3 and solve for A:

8x – 10 = A(2x + 1) + B(x – 3)

8(3) – 10 = A(2(3) + 1) + B(3 – 3)

14 = 7A

2 = A

Now set x = 0 and solve for B:

8x – 10 = A(2x + 1) + B(x – 3)

8(0) – 10 = (2)(2(0) + 1) + B(0 – 3)

-10 = 2 – 3B

3B = 12

B = 4

The decomposition is

Example 11

Decompose into partial fractions.

Answer

The denominator factors as

-2x2 + 7x + 15 = (2x + 3)(5 – x)

or as

-2x2 + 7x + 15 = (-2x – 3)(x – 5).

We'll go with

-2x2 + 7x + 15 = (2x + 3)(5 – x)

first. Then the decomposition will look like

Add the partial fractions and set the resulting numerator equal to the original numerator:

-18x – 1 = A(5 – x) + B(2x + 3).

Take x = 5 and solve for B:

-18x – 1 = A(5 – x) + B(2x + 3)

-18(5) – 1 = A(5 – 5) + B(2(5) + 3)

-91 = 13B

-7 = B

Take x = 0 and solve for A:

-18x – 1 = A(5 – x) + B(2x + 3)

-18(0) – 1 = A(5 – 0) + (-7)(2(0) + 3)

-1 = 5A – 21

4 = A

The decomposition is

If instead we had started by factoring the denominator as

-2x2 + 7x + 15 = (-2x – 3)(x – 5)

then the decomposition would look like

Add the partial fractions and set the resulting numerator equal to the original numerator:

-18x – 1 = A(x – 5) + B(-2x – 3).

Set x = 5 and solve for B:

-18x – 1 = A(x – 5) + B(-2x – 3)

-18(5) – 1 = A(5 – 5) + B(-2(5) – 3)

-91 = -13B

7 = B

Set x = 0 and solve for A:

-18x – 1 = A(x – 5) + B(-2x – 3)

-18(0) – 1 = A(0 – 5) + 7(-2(0) – 3)

-1 = -5A – 21

20 = -5A

-4 = A

Then the decomposition looks like this:

While these two answers may look different, they are in fact the same. If we multiply each term of our first answer by 1, we get

which is the same as our second answer.

Example 12

Decompose into partial fractions.

Answer

The denominator factors nicely into

2x2 + 14x = 2x(x + 7)

so the decomposition will look like

Add the partial fractions and set the resulting numerator equal to the original numerator:

15x + 21 = A(x + 7) + B(2x).

Set x = 0 and find A:

15x + 21 = A(x + 7) + B(2x)

15(0) + 21 = A(0 + 7) + B(2(0))

21 = 7A

3 = A

Set x = 1 and find B:

15x + 21 = A(x + 7) + B(2x)

15(1) + 21 = (3)(1 + 7) + B(2(1))

36 = 24 + 2B

6 = B

The decomposition is

Example 13

Decompose into partial fractions.

Answer

The denominator factors into

3x2 + x – 4 = (x – 1)(3x + 4)

so the decomposition into partial fractions will look like

Add the partial fractions and set the resulting numerator equal to the original numerator:

25x + 10 = A(3x + 4) + B(x – 1).

Set x = 1 and solve for A:

25x + 10 = A(3x + 4) + B(x – 1)

25(1) + 10 = A(3(1) + 4) + B(1 – 1)

35 = 7A

5 = A

Set x = 0 and solve for B:

25x + 10 = A(3x + 4) + B(x – 1)

25(0) + 10 = (5)(3(0) + 4) + B(0 – 1)

10 = 20 – B

10 = B

The decomposition is

Example 14

Integrate.

Answer

So

Example 15

Integrate.

Answer

This means

Example 16

Integrate.

Answer

So

Example 17

Integrate.

Answer

We didn't do this one yet. The denominator factors into

-2x2 + 3x – 1 = (-2x + 1)(x – 1)

so the partial fraction decomposition will look like

Add the partial fractions and set the resulting numerator equal to the original numerator:

-3x + 1 = A(x – 1) + B(-2x + 1).

Set x = 1 and solve for B:

-3x + 1 = A(x – 1) + B(-2x + 1)

-3(1) + 1 = A(1 – 1) + B(-2(1) + 1)

-2 = -B

2 = B

Then set x = 0 and solve for A:

-3x + 1 = A(x – 1) + B(-2x + 1)

-3(0) + 1 = A(0 – 1) + (2)(-2(0) + 1)

1 = -A + 2

A = 1

The partial fraction decomposition is

Now we can evaluate the integral:

Example 18

Integrate.

Answer

The denominator factors as

6x2 + 5x + 1 = (2x + 1)(3x + 1)

so the partial fraction decomposition will look like

Add the partial fractions and set the resulting numerator equal to the original numerator: